Question 887075
Use Descartes Rule of Signs to state the number of possible positive real zeros and the number of negative real zeros of P(x)=3x^4 - 6x^3 + 2x^2 - 7x - 4
 Please show the work so I can see how the problem was worked out. Thanks
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P(x)=3x^4 - 6x^3 + 2x^2 - 7x - 4
# of changes of sign in P(x) = 3
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Using Descarte's Rule::
# of Real Roots <= # of sign changes in P(x) - 0 or 2
For P(x) that means # of Real Roots in P(x) is 3 or 1
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P(-x) = 3x^4 + 6x^3 + 2x^2 + 7x - 4
# of sign changes in P(-x) = 1
Using Descarte's Rule::
# of Real Roots <= # of sign changes in P(-x) - 0
For P(-x) that means # of Real Roots in P(x) is 1
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Conclusion for your problem::
# of Real Roots is 1
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For additional narrative and examples using Descarte's Rule
Google "Descartes Rule",
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Cheers,
Stan H.